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Multiple Choice questions on Electric current resistance and resistivity





Assertion and Reason Type Question

Following question consider two statements one labelled as statement-1 and another as statement-2. Study both statements and mark your answer using the codes given blow.
  1. Both Statement-1 and Statement-2 are true and Statement-2 is correct explanation of Statement-1.
  2. Both Statement-1 and Statement-2 are true and Statement-2 is not a correct explanation of Statement-1.
  3. Statement-1 is true but statement-2 is false.
  4. Statement-1 is false and statement-2 is true.
Question-1
Statement-1 Free electrons in conductor are always in state of continuous random motion.
Statement-2 The random motion of free electrons is due to thermal energy of the conductor.

Question-2
Statement-1 Drift velocity of electrons in a conductor increase on increasing the temperature of the conductor.
Statement-2 On increasing the temperature of a conductor, resistivity of the conductor increases.

Question-3
Statement-1 For metals relaxation time for electrons decreases with increase in temperature.
Statement-2 With the increase in temperature, number of collision per unit time that electrons made with lattice ion also increases.

Answer

(1) a
(2) d
(3) b


Multiple Choice Type Questions


Question-4
Resistivity of the material of a conductor having uniform area of cross-section varies along its length according to the relation
ρ= ρ0 (a+bx)
if L is the length of the conductor and be the area of cross-section then resistivity of the conductor given by relation
Multiple Choice questions on Electric current resistance and resistivity for Jee Main and Advanced

Answer

Consider an infinitesimally small length element dx of the conductor at a distance x from its one end. Then, the resistance dR of the length element dx of the conductor is
$dR=\rho \frac {dx}{A}= \rho _0 (a+bx) \frac {dx}{A}$
The resistance of the whole length of the conductor is
$R=\int_{0}^{L} dR= \int_{0}^{L} \rho _0 (a+bx) \frac {dx}{A} =\frac {\rho _0 }{A} (aL + \frac {bL^2}{2})$
obtained on solving the integral


Question-5
Given a current carrying wire of non-uniform cross-section. Which of the following is constant through out the wire.
(a)Current only
(b)Current and drift speed
(c)Drift speed only
(d)Current, drift speed and electric field

Answer

We know that I = neAvd , Drift Velocity$v_d=\frac {1}{neA} \alpha \frac {1}{A}$ ,
For non-uniform cross-section drift speed will be different for different situations. Only current (rate of flow of charge) will same.


Question-6
When potential difference across a given copper wire is increase, drift velocity of charge carriers
(a)Decreases
(b)Increases
(c)Remain same
(d)Get reduced to zero

Answer

We know that
$v_d=\frac {eE}{m} \tau $
If L is the length of the copper wire and V is the potential difference across it then
$v_d=\frac {e}{m} \frac {V}{L} \tau $
$v_d \alpha V$ i.e. it potential difference is increase drift velocity of electrons also increases



Question-7
A material B has twice the specific resistance of the material A. A circular wire made of B has twice the diameter of the wire made of A. Then , for the two wires to have to have the same resistance, the ratio
La/Lb of their respective lengths must be
(a) ½
(b) 2
(c) ¼
(d)1

Answer

Let La and Lb be the lengths of two wires.
Given that ρA=ρ i.e. ρB=2ρ also diameter of wire A is DA=D then for wire B DB=2D,
The resistance of the wire made of material A,
$R_A= \frac {\rho _A L_A}{\pi (\frac {D_A}{2})^2} = \frac {\rho _A L_A}{\pi (\frac {D}{2})^2} = \frac {4\rho L_A}{\pi D^2}$
Resistance of the wire made of material B
$R_B= \frac {\rho _B L_B}{\pi (\frac {D_B}{2})^2} = \frac {\rho _B L_B}{\pi (\frac {2D}{2})^2} = \frac {2\rho L_B}{\pi D^2}$
Since RA=RB, we have
$\frac {4\rho L_A}{\pi D^2}=\frac {2\rho L_B}{\pi D^2}$
$\frac {L_A}{L_B} =\frac {1}{2} $


Question-8
What will be the equivalent resistance between A and D
Multiple Choice questions on Electric current resistance and resistivity for Jee Main and Advanced
(a)40 Ω
(b)20Ω
(c)30Ω
(d)10Ω

Answer

Resistance at the end B and C do not be the part of network of resistance between point A and D.
Therefore equivalent resistance between points A and D is
$R= 10 + \frac {1}{10+10} + \frac {1}{10+10} +10$
$R=30 \Omega $



Question-9
In the given circuit below the equivalent resistance between points A and C
Multiple Choice questions on Electric current resistance and resistivity for Jee Main and Advanced
(a)32
(b)30.7
(c)33.07
(d)3.07

Answer

Resistance of the segment AB and BC are in series. Therefore effective resistance is
$R_1=9+5=14\Omega$
Similarly resistance of the segment AD and DC are in series. IfR2 is the effective resistance
$R_2 =3+6 =9 \Omega$
The resistanceR1 andR2 are in parallel with $7 \Omega $ resistor. Then equivalent resistance is
$\frac {1}{R_eq}= \frac {1}{14} + \frac {1}{9} + \frac {1}{7]$
Req =3.07 $\Omega$


Question-10
A circular ring having negligible resistance is used to connect four resistors as shown below in the figure.
The equivalent resistance between point A and B is
Multiple Choice questions on Electric current resistance and resistivity for Jee Main and Advanced
(a)10R
(b)2R
(c)3R
(d)R

Answer

Above network of resistors is equivalent to the arrangement shown below in the figure
Multiple Choice questions on Electric current resistance and resistivity for Jee Main and Advanced
If R1 is equivalent resistance of parallel combination then
$\frac {1}{R_1 = \frac {1} {3R} + \frac {1} {3R} + \frac {1} {3R}$
R1 = R
Therefore equivalent resistance of the network between A and B is
Req=R+R=2R


Paragraph Type Questions

Two concentric metal spherical shells of radius a and b respectively are separated by weakly conducting material of conductivity σ
A potential difference V is applied across the two shells
Question 11
Find the resistance between the shells
Practice Questions on Electric current resistance and resistivity for JEE Main and Advanced

Answer

Consider a shell of radius r and thickness dr as shown in figure

Resistance of the shell
$dR=\frac{dr}{\sigma4\pi r^2}$

Total resistance

$R=\int_{a}^{b}\frac{dr}{\sigma4\pi r^2}$
$=\frac{1}{4\pi\sigma}(\frac{1}{a}-\frac{1}{b})$


Question 12
Find the electric field at a distance r from the center
Practice Questions on Electric current resistance and resistivity for JEE Main and Advanced

Answer

We know that
$J=\frac{I}{A}$

And
$J=\sigma E$

So we can say that
$E=\frac{I}{\sigma A}$
Now
$V=IR$
So
$E=\frac{V}{\sigma RA}$
For distance r from center
$E=\frac{V}{\sigma R4\pi r^2}$
Substituting the value of R from above solution
$E=\frac{V}{(\frac{1}{a}-\frac{1}{b})}\frac{1}{r^2}$


Multiple Choice Type Questions


Question 13
A potential difference of V is applied at the ends of a copper wire of length l and diameter d. On doubling only d, drift velocity
(a) Becomes two times
(b) Becomes half
(c) Does not change
(d) Becomes one fourth

Answer

Drift velocity is given by
$v_d=\frac {V}{\rho l ne}$
$v_d$ does not depend upon diameter.
Hence (c) is the correct option


Question 14
The effective resistance between points P and Q of the electrical circuit shown in the figure is

(a) $\frac {2Rr}{R+r}$
(b) $8R(R+r)(3R+r)$
(c) $2r+4R$
(d) $\frac {5R}{2} + 2r$

Answer

In a circuit, any circuit element placed between points at the same potential can be removed, without affecting the rest of the circuit, here, by symmetry, points P, Q and R are at same potential, for any potential difference between A and B

The circuit can therefore be reduced as shown below

Now this can be reduced to

Therefore Effective Resistance
$\frac {1}{R_{eq}} = \frac {1}{4R} + \frac {1}{4R} + \frac {1}{2r}$
$\frac {1}{R_{eq}} =\frac {r + r + 2R}{4Rr}$
$R_{eq}=\frac {2Rr}{R+r}$


Question 15
Which of the following characteristics of electrons determines the current in a conductor?
(a) Drift velocity alone.
(b) Thermal velocity alone.
(c) Both drift velocity and thermal velocity.
(d) Neither drift nor thermal velocity.

Answer

(a)


Question 16
Water of volume 2 litre in a container is heated with a coil of 1 kW at 27 °C. The lid of the container is open and energy dissipates at rate of 160 J/s. In how much time temperature will rise from 27°C. to 77°C. [Given specific heat of water is 4.2 kJ/kg]
(a) 8 min 20 s
(b) 6 min 2 s
(c) 7 min
(d) 14 min

Answer

Heat gained by water = Heat supplied by container - heat lost
$m s \Delta T = 1000t - 160t $
$t= \frac {2 \times 4.2 \times 1000 \time 50}{840}$ = 8 min 20 sec


Question 17
If a wire is stretched to make it 0.1% longer, its resistance will
(a) increase by 0.2%
(b) increase by 0.05%
(c) decrease by 0.2%
(d) decrease by 0.05%

Answer

(a)


Question 18
Six identical resistors are connected as shown in the figure. The equivalent resistance will be

(a)Maximum between P and R
(b)Maximum between Q and R
(c)Maximum between P and Q
(d) All are equal

Answer

(c)


Question 19
An ideal gas is filled in a closed rigid and thermally insulated container . A coil of 100 ω resistor carrying a current of 1 A for 5 minutes supplies heat to the gas . The change in the internal energy of the gas is
(a) 10 kJ
(b)30 kJ
(c) 20 kJ
(d) 0 kJ

Answer

Here,heat supplied = the change in internal energy
$Q= \Delta U$
Now heat supplied $Q=i^2 R t$
$=1 \times 1 \times 100 \times 5 \times 60=30000$ J=30kJ.
Hence (b) is the correct option


Question 20
Temperature dependence of resistivity $\rho(T)$ of semiconductors, insulators and metals is significantly based on the following factors:
(a) number of charge carriers can change with temperature T.
(b) time interval between two successive collisions can depend on T.
(c) length of material can be a function of T.
(d) mass of carriers is a function of T.

Answer

(a), (b)











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